I would please like some help on the exercise 257 of the Rose Group Theory. Let $G$ be a finite group and $H$ and $K$ be normal subgroups of $G$, and $P$ a Sylow p-subgroup of $G$. Then $(PH)\cap (PK)=P(H\cap K)$.
One inclusion is straightforward: $P(H\cap K)\subset (PH)\cap (PK)$ since $H\cap K \subset H,K$. Let $g\in G$ such that $g=p_1h=p_2k$ where $p_i \in P, h\in H,k\in K$. I want to show $p_1=p_2$. The normality of H and K tells me that PH and PK are subgroups of G. By order considerations, P is a p-Sylow of PH, PK and $P(H\cap K$). By normality, $H \triangleleft PH, K \triangleleft PK,H\cap K \triangleleft P(H\cap K)$. I thought about the Frattini's argument but nothing there.
Notice that $H\cap K$ is also a normal subgroup of $G$. Let $\bar G =G/H\cap K$. Notice also that $\bar P$ is also a sylow $p$ subgroup of $\bar G$.
$\bar P \bar H \cap \bar P \bar K=\bar P (\bar H\cap \bar P \bar K)$ by Dedekind rule.
Let $h\in \bar H\cap \bar P \bar K$ then $h=ak$ for $a\in \bar P,k\in \bar K$. Note that the elements $h,k$ commutes with each other as $\bar H\cap \bar K=1$. Hence
$$(hk^{-1})^{|a|}=a^{|a|}=e$$ $$h^{|a|}=k^{|a|}\implies h^{|a|}=e$$
Hence $h$ is element of prime power order. Thus, $ \bar H\cap \bar P \bar K$ is an $p$ group $\implies \bar P (\bar H\cap \bar P \bar K)$ is an $p$ group containing $\bar P$. By the maximality of $\bar P$, we have
$$\bar P \bar H \cap \bar P \bar K=\bar P$$ Hence $$PH\cap PK=P(H\cap K)$$
Note For Dedekind Rule, check here.